LED board design troubles

Thread Starter

bitzer

Joined Aug 26, 2011
3
gidday all! first post (of what i am sure will be many) on this here forum! i have recently been attempting to design a LED light board for camping as a 'camp' light. the power source is a 12v car battery which also powers my portable fridge. now, the design i plan on is 60 LED's, hooked up in 2 groups of 3 rows of 10...if that makes sense. 10 bulbs in series, total of 6 rows but in 3 row groups. Bulbs are rated as a forward voltage of 3.2v @ 20mA, my questions are:
1) will 10 bulbs in series mean that the last bulb is dimmer than the first?
2) are my calculations correct when i say that the 3 rows of LED's will draw 3.2v each, thus a total of 9.6v per group is used. 12v - 9.6=2.4v 2.4V x total of 200mA = 480Ω resistor needed per group...right??

please excuse any 'noobness' in above post :D i have VERY little idea but i am trying to learn!! this project may seem impractical due to cost but i just want to have a go at building a light circuit. i can solder....sort of! enough to join wires...this project is gonna be a little fun i think!!

Cheers, Jake
 

Wendy

Joined Mar 24, 2008
23,415
I think I know your problem right off.

LEDs are current devices, but they also drop a set voltage. White LEDs typically drop 3.6 volts each, so if you have 10 in series as you have suggested in the first paragraph it would need 36 V, actually more.

Try 3 LEDs per chain, which will add up to 10.8V. If the power supply is a stable 12V (no guarantees with a car battery) you will need approximately 450Ω. This can be rounded off to a standard value of 470Ω.

Beginners are always welcome here, we have a firm no flame policy from folks who consider themselves more advanced. Questions are always welcomed.

I've written a tutorial you might like...

LEDs, 555s, Flashers, and Light Chasers
 

SgtWookie

Joined Jul 17, 2007
22,230
Just so you know, automotive batteries are not designed to be used in the manner you are considering using them. They will rapidly fall apart if deeply discharged, even if only a few times.

Automotive batteries are designed to give out brief (30 seconds max) periods of very high current (up to 500 Amperes) in order to start an engine, and then be immediately recharged by the engines' alternator. They can last for many years when used in this manner.

You will need what is known as a "deep cycle" battery. Batteries designed for marine use (boating) are usually a compromise between deep-cycle and automotive batteries. They will last much longer in deep-cycle use than an auto battery, but not as long as a true deep-cycle battery.

Refrigerators are usually very power-hungry, and will probably drain your battery quite quickly. You are usually much better off to use large blocks of ice rather than try to use electricity. Blocks of ice are much cheaper than large batteries.
 

iONic

Joined Nov 16, 2007
1,662
Without regulating the voltage to the LED's, each time your refrigerator kicks on you will know it as the LED's will likely dim.

@Bill,
I'm trying to figure out where you came up with 450 Ohm resistors. Perhaps using a voltage drop of 3.6V/LED. The OP mentioned 3.2V. But that actually shrinks the needed resistor.

Here are my calculations:

Source voltage: 12.4V(as we know an auto battery is not 12V, it actually should be charging at this point, especially with SLA and the OPs application. Technically since he should be charging by 12V, before really, I just chose a value between the nominal 12.7V and 12V.)
LED forward Voltage: 3.2V
LED forward Current 18mA(Safety factor for longevity)

Thus: 3(3.2V) = 9.6V
Remaining Voltage: 12.4V - 9.6V = 2.8V
Desired current: 18mA
Series resistance: 2.8V/.018A = 156 Ohms (150 Ohms will do)

How are we a factor of three off?
 

Thread Starter

bitzer

Joined Aug 26, 2011
3
wow! thanks for all the advice and info! couple things need clarifying but :D
@Bill, so youre saying to use (per group) 10 rows of 3 bulbs wired in series instead of 3 rows of 10 in series....right??

@SGTWOOKIE, sorry, i should have mentioned that yes it is a deep cycle. i have a dual battery system in the ute which charges it while on the move. the fridge is a 3-way fridge meaning it runs off mains, 12v and gas. i use gas when i am camped (provided i have some!!) but run it off the battery while driving and on short stops. occasionally i will be running both light and fridge but only when ive got no gas.

@iONic, so ive kind of worked out that unless the bulbs get a minimum of 3.2v, they wont glow..now, the amps. these bulbs are rated at 0.20mA, yet you state "0.18mA for longevity" so if they run on 0.18mA they will be dimmer, but last longer? so if that is the case, then its the amps that determine how bright a bulb will be (provided you stay below the rating)

one last thing...say i want to wire in a variable resistor (POT..?) to dim the light to required brightness..would i build the circuit, with 10 rows of 3 bulbs in series, with each row having a 150ohm resistor, then simply 'wire in' the POT on the main wiring between the power supply and resistors???

i greatly appreciate the time and patience people are taking with me!!
cheers, Jake
 

iONic

Joined Nov 16, 2007
1,662
wow! thanks for all the advice and info! couple things need clarifying but :D
@Bill, so youre saying to use (per group) 10 rows of 3 bulbs wired in series instead of 3 rows of 10 in series....right??
Yes, 3 rows of 10 in series will require more than 30V.

@iONic, so ive kind of worked out that unless the bulbs get a minimum of 3.2v, they wont glow..now, the amps. these bulbs are rated at 0.20mA, yet you state "0.18mA for longevity" so if they run on 0.18mA they will be dimmer, but last longer? so if that is the case, then its the amps that determine how bright a bulb will be (provided you stay below the rating)
Not exactly, they will glow with less voltage, but the brightness can not be easily varied by adjusting the voltage. It's possible that once you reach 2.5V/LED the light will be pretty much worthless. Current is the best way to vary the brightness of LED's.
For LED life, it will be prudent to stay slightly under the typical/nominal rated current. Each LED has it tolerances for rated voltage and current, just like resistors... 1%, 2%, 5%...etc. The visual difference between 18ma and 20mA will likely not be noticeable.

Note: in the schematic below I have revised my resistor value to 180 ohms. It will do a better job of keeping the LED current 20mA or less when your battery is at 12.7V and slightly reduce the current when the battery is at a lower voltage.



one last thing...say i want to wire in a variable resistor (POT..?) to dim the light to required brightness..would i build the circuit, with 10 rows of 3 bulbs in series, with each row having a 150ohm resistor, then simply 'wire in' the POT on the main wiring between the power supply and resistors???

i greatly appreciate the time and patience people are taking with me!!
cheers, Jake
Here is one of several ways to achieve different brightnesses
Use three switches, SW1 as both main power and to power 4 legs of the 10
sw2 will add three more legs, 9 more LED's, equaling 7 legs
sw3 will add three more legs, 9 more LED's, equaling 10 legs

It doesn't matter which switch you use sw2 or sw3 in whatever order, the outcome is the same, except when sw1 is off, then no light is possible.




Another Method:



This method used PWM to achieve a variable brightness with just one Pot.
 
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Thread Starter

bitzer

Joined Aug 26, 2011
3
so what would happen if i wanted 2 modes of brightness? could i hook up the groups with 240ohm resistors, meaning that the LED's are running on 3.2v (working around 12v) @ 10mA, making them dimmer, than using a 3 way switch, add another leg to the circuit before the LED's using 300ohm resistors to provide another 8mA bringing the total power to LEDs to 18mA...would there be a significant variation in light output? would it even work?
 

iONic

Joined Nov 16, 2007
1,662
so what would happen if i wanted 2 modes of brightness? could i hook up the groups with 240ohm resistors, meaning that the LED's are running on 3.2v (working around 12v) @ 10mA, making them dimmer, than using a 3 way switch, add another leg to the circuit before the LED's using 300ohm resistors to provide another 8mA bringing the total power to LEDs to 18mA...would there be a significant variation in light output? would it even work?
You really got me confused now...
Yes you could replace all the 180 Ohm resistors with 240 Ohm resistors which would make your whole light array dimmer right from the start. But if you are thinking that adding another whole array or 30 LEDs, 3 to each leg prior to(is series) with the other array using 300 ohm resistors will add the current 10mA + 8mA, not that is not a correct assumption. I wouldn't recommend limiting your LED's to 10mA with a resistor as It will waste more power in the resistors than your using with the LED's. The PWM method would best serve you in that regard. If you only want 2 brightnesses, then you could use a position toggle switch(On1-Off-On2). Center would be off, On1 - Full Brightness, and On2 - Reduced Brightness.

30 high brightness LED's are not bad for fairly close up stuff such as reading or general lighting at a little greater distance. I used to have a small reading light to read by at night that used 30 LED's and ran off of 4.5V. It wasn't too bad, but I replaced the 30 LED's with 1 High Power LED and added the PWM circuit for dimming and it is even brighter than before. It's a single 120 degree LED that runs off of 12V and 350mA.

 
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